I did that now ... The voltage across the two parallel resistors is 0.192 V so I guess if the I = V / R then 0.192 / 5 = 0.0384 Amps ?

Is that normal or ?

 

Yes, 38mA to energize the relay is no surprise. It doesn't answer why the transformer is getting hot.

 

Ummmmm

Ok ok I will confess

When I first time used the transformer I connected the secondary to the "240 V" side by mistake

Of course this burned two fuses in the plug

Those chinese doesn't mention where is the LV and HV

But after that it gives the 12 V and everything ...

Would that be a reason ?? :'(

 

Hello,

An other thing that worries me about the schematic is the use of the 555 in this schematic.
If you use an 15 Volts transformer, the voltage on the 555 will be about 20 volts.
In ALL datasheets of the 555 there is an ABOLUTE maximum powersupply of 18 Volts given.
The way the 555 is used now, it probably won't last long.

Bertus

 

Hello,

An other thing that worries me about the schematic is the use of the 555 in this schematic.
If you use an 15 Volts transformer, the voltage on the 555 will be about 20 volts.
In ALL datasheets of the 555 there is an ABOLUTE maximum powersupply of 18 Volts given.
The way the 555 is used now, it probably won't last long.

Bertus

Thank you Bertus I really appreciate it. In these very technical stuff, I have a zero knowledge so I guess you are right. I think like I said before I will reconsider building another more effective circuit .. If just I can find a good transformer with higher ratings.

In such case would you suggest replacing 555 with another IC ?

 

Hello,

Have a look at the circuit that SgtWookie made in post #4 of the following thread:
http://forum.allaboutcircuits.com/showthread.php?t=62386

It is designed for smaller batteries but it might give you some idea.

Bertus

 

Hello,

Have a look at the circuit that SgtWookie made in post #4 of the following thread:
http://forum.allaboutcircuits.com/showthread.php?t=62386

It is designed for smaller batteries but it might give you some idea.

Bertus

Thank you very much Bertus .. I will check it ... I will may get back to here or there is I need anything ... Thank you very much .... and all the guys ... <3 <3 <3

 

I think you should initially limit your circuit to the transformer , rectifier and smoothing capacitor and see what current it can safely deliver without overheating the transformer . You will not damage a car type battery (30-40 Ah) if the charging voltage < 18volts for a short period (15-20 mins).

The 6A4 Diode on the positive rail between the rectifier and the rest of the circuit serves no useful purpose and will cause 0.7V drop in the output voltage. It may be deliberately placed there solely for that reason to limit the voltage. The two output resistors have a combined resistance of 5 Ohm. This value is far too high it should be more like 0.5 Ohm. 5 Ohm @ 1 A = 5Volts drop.
I would agree that a fuse or thermal trip is a must.
Finally may I suggest that you consider modifying your circuit to prevent the relay from closing unless the polarity of the battery is correct. It is a very common problem for the battery leads to be accidently reversed.
Finally your charge current is largely dependant on the state of charge of the battery i.e. if the battery is near full charge then current will be low.

 

All the charging current has to pass through the paralleled 10 ohm resistors.. ie 5 ohms. In that particular case the short circuit current you measured was actually safe because there is a load - 5 ohms. That's why you saw 1.5 amps.
I can't see this being anything other than a float charger. Even 1 amp current will cause a 5 v drop across that series resistance. and 5 amp an impossible 25v !
You need a minimum of 13.8 dc across a 12v lead acid battery to make any sort of charger and at that voltage it will take several days to reach 100% charge. At 14.2 dc charging would be quicker but even with a 20v ac transformer and 10 times bigger reservoir capacitor the current may be limited to about 2 amps. Anyway,t hose Resistors would be at max dissipation at 2 amps.
Was there any text with the schematic you found that gave a clue about it's operation and spec. ?

edit: cork ie posted at the same time !..and I missed a whole page..yes 5 ohm, ,He maybe misread the value. 0.5 quite possibly.

 

You can find a lot of bad circuits ideas on the net. And I think this is one of them.

 

I think you should initially limit your circuit to the transformer , rectifier and smoothing capacitor and see what current it can safely deliver without overheating the transformer . You will not damage a car type battery (30-40 Ah) if the charging voltage < 18volts for a short period (15-20 mins).

I guess I should do that .. But it will take time to melt the solder .. etc

The 6A4 Diode on the positive rail between the rectifier and the rest of the circuit serves no useful purpose and will cause 0.7V drop in the output voltage.

You are absolutely right ... I bypassed that diode with a wire like wayneh said .. both supply voltage and current increased a little.

The two output resistors have a combined resistance of 5 Ohm. This value is far too high it should be more like 0.5 Ohm. 5 Ohm @ 1 A = 5Volts drop.

I wish the passing ampere rich that value .. right now it's just 0.2 amp (200mA)

I would agree that a fuse or thermal trip is a must.

Yeah I will man )

Finally may I suggest that you consider modifying your circuit to prevent the relay from closing unless the polarity of the battery is correct. It is a very common problem for the battery leads to be accidently reversed.

Well, to be honest I can't design circuits, I just build what's already designed so this will be a little hard for me right now

Finally your charge current is largely dependant on the state of charge of the battery i.e. if the battery is near full charge then current will be low.

Well I guarantee that the battery is not charged at all, both hydrometer and the car says that


Thank you very much I appreciate your reply and help <3

 

All the charging current has to pass through the paralleled 10 ohm resistors.. ie 5 ohms. In that particular case the short circuit current you measured was actually safe because there is a load - 5 ohms. That's why you saw 1.5 amps.
I can't see this being anything other than a float charger. Even 1 amp current will cause a 5 v drop across that series resistance. and 5 amp an impossible 25v !
You need a minimum of 13.8 dc across a 12v lead acid battery to make any sort of charger and at that voltage it will take several days to reach 100% charge. At 14.2 dc charging would be quicker but even with a 20v ac transformer and 10 times bigger reservoir capacitor the current may be limited to about 2 amps. Anyway,t hose Resistors would be at max dissipation at 2 amps.
Was there any text with the schematic you found that gave a clue about it's operation and spec. ?

edit: cork ie posted at the same time !..and I missed a whole page..yes 5 ohm, ,He maybe misread the value. 0.5 quite possibly.

Thanks cornishlad ... Look at the attachments

This is the output voltage from the circuit >>

The current still 0.2 amp ... The battery is not charged ... At least I will save days if I can make the current reaches 1 A (

I don't know if I can get a link for the circuit site .... but I may quote the operation that written for now >>>

This is a simple DIY charge controller schematic posted in response to a request by one of our readers on our facebook page. The main component of this auto battery charger circuit is a 555 timer which compares the voltage in the battery. It turns ON the charger if the battery voltage is below the variable preset voltage (12 volt chosen here) and turns OFF the charger if the voltage reaches 13.8 volt. The battery charging voltage of the charger can be varied by adjusting the variable resistor and maximum charging is limited by a 13.8V zener diode on the fifth terminal of 555 IC.

Working

Positive terminal of the upper comparator of 555 is connected with 13.8V in order to turn OFF the charger if the battery charges above 13.8V.
13.8V is obtained by connecting a 13.8V zener in series with a resistor.
If the battery voltage is greater than 13.8V, comparator output goes high and flip flop will be set. This turns OFF the transistor and the relay. For understanding relay working see How to connect relay: Relays working with animation
If the battery voltage is below the preset voltage (set by us), lower comparator will reset the flip flop. This turns ON the transistor and the relay will switch to charge the battery.
The recharge voltage (preset voltage) can set by varying the variable resistor.
Power ON is indicated by a red LED and charger ON status is indicated by a green LED.

That's all what is written ... with components and diagrams .. etc

 

If you want a car battery charger. Get one those simple impulse charges. That is just a transformer and a rectifier. It is really hard to overcharge a car car battery with one of those. But as you may now car batteries do not last forever. They do have a life span.

 

You can find a lot of bad circuits ideas on the net. And I think this is one of them.

I agree with you ... I tried a lot of it !!

But ... Can you give me other worked ones ? )

 

If you want a car battery charger. Get one those simple impulse charges. That is just a transformer and a rectifier. It is really hard to overcharge a car car battery with one of those. But as you may now car batteries do not last forever. They do have a life span.

Yeah of course ... But you know people here doesn't have any ideas about charging a car battery ... they just replace after it's first cycle !!!

This is the case now ... this is it's first cycle ... i'm sure it can be charged but with reasonable circuit (device)

 

OK guys Help

Is it a charging or discharging circuit !!

The voltage of the battery have dropped to 10.9 V !!!

Good news is the current have increased to 0.3 A LOL

 

As several others have suggested. ditch the whole 555 circuit and connect the rectifier, optionally with reservoir cap, to the battery. with a 12 or 13v ac transformer it will work as a "taper charger" and eventually charge the battery.

 

There's no reason (other than saving 38mA) to ditch the 555 part of the circuit; it's not making any difference. The relay is always on.

 

What should I do now

 

In this photograph:

the battery case appears to be bulging, particularly on the left end. This indicates that the cells are damaged, and the battery needs to be replaced.